Formula list

Fx = F cos θ · Fy = F sin θ

F

magnitude of the vector

θ

angle to the chosen axis° or rad

The syllabus requires a vector to be represented as two perpendicular components, and coplanar vectors to be added and subtracted.

s = ut + ½at²

s

displacementm

u

initial velocitym s⁻¹

a

accelerationm s⁻²

t

times

v² = u² + 2as

v

final velocitym s⁻¹

u

initial velocitym s⁻¹

a

accelerationm s⁻²

s

displacementm

v = u + at

v

final velocitym s⁻¹

u

initial velocitym s⁻¹

a

accelerationm s⁻²

t

times

Not on the data sheet. Two of the four equations of motion are printed for you and two are not — this is one of the two to memorise.

s = ½(u + v)t

s

displacementm

u, v

initial and final velocitym s⁻¹

t

times

Not on the data sheet.

velocity = gradient of s–t · acceleration = gradient of v–t · displacement = area under v–t

gradient

of a tangent for non-uniform motion

F = ma

F

resultant forceN

m

masskg

a

accelerationm s⁻²

Force and acceleration are in the same direction. A special case of F = Δp/Δt for constant mass.

p = mv

p

momentumkg m s⁻¹, N s

m

masskg

v

velocitym s⁻¹

F = Δp / Δt

F

resultant forceN

Δp

change in momentumkg m s⁻¹

Δt

time takens

total momentum before = total momentum after

p

summed over the system, with direction

For a closed system. Applies in one and two dimensions.

relative speed of approach = relative speed of separation

v

relative speed along the line of collisionm s⁻¹

The test for a perfectly elastic collision; total kinetic energy is also conserved.

moment = F × d

F

forceN

d

perpendicular distance from the pivotm

torque = F × d

F

one of the two equal and opposite forcesN

d

perpendicular separation of the forcesm

resultant force = 0 · resultant moment = 0

about

any point, for the moment condition

The principle of moments, plus a closed vector triangle for three coplanar forces.

ρ = m / V

ρ

densitykg m⁻³

m

masskg

V

volumem³

p = F / A

p

pressurePa

F

force normal to the surfaceN

A

aream²

p = ρgh

p

pressure due to the columnPa

ρ

density of the fluidkg m⁻³

g

acceleration of free fallm s⁻²

h

depthm

The syllabus also asks you to derive it from the definitions of pressure and density.

F = ρgV

F

upthrustN

ρ

density of the fluidkg m⁻³

V

volume of fluid displacedm³

W = Fs cos θ

W

work doneJ

F

forceN

s

displacementm

θ

angle between force and displacement°

Work is force × displacement in the direction of the force.

P = W / t

P

powerW

W

work doneJ

t

times

P = Fv

P

powerW

F

forceN

v

velocitym s⁻¹

The syllabus asks you to derive this one.

efficiency = useful energy output / total energy input

energy

or power, in the same unit top and bottom

EP = mgh

EP

change in g.p.e.J

m

masskg

g

gravitational field strengthN kg⁻¹

h

change in heightm

For a uniform field. Derived from W = Fs.

EK = ½mv²

EK

kinetic energyJ

m

masskg

v

speedm s⁻¹

The syllabus asks you to derive this from the equations of motion.

F = kx

Rearrangedk = F / x · x = F / k

F

loadN

k

spring constantN m⁻¹

x

extensionm

σ = F / A

σ

stressPa

F

forceN

A

cross-sectional aream²

ε = x / L

ε

strain (a ratio, no unit)

x

extensionm

L

original lengthm

E = σ / ε

E

Young modulusPa

σ

stressPa

ε

strain

EP = ½Fx = ½kx²

EP

strain energyJ

F

loadN

x

extensionm

k

spring constantN m⁻¹

Within the limit of proportionality. It is the area under a force–extension graph.

v = fλ

Rearrangedf = v / λ · λ = v / f

v

wave speedm s⁻¹

f

frequencyHz

λ

wavelengthm

The syllabus asks you to derive it from the definitions of speed, frequency and wavelength.

intensity = power / area

I

intensityW m⁻²

P

powerW

A

aream²

intensity ∝ (amplitude)²

I ∝ A²

for a progressive wave

fo = fs v / (v ± vs)

fo

observed frequencyHz

fs

source frequencyHz

v

speed of soundm s⁻¹

vs

speed of the sourcem s⁻¹

Minus when the source approaches, plus when it recedes.

I = I₀ cos²θ

I

transmitted intensityW m⁻²

I₀

intensity of the plane-polarised waveW m⁻²

θ

angle between the polariser axes°

λ = ax / D

λ

wavelengthm

a

slit separationm

x

fringe separationm

D

slit-to-screen distancem

For light, with D much greater than a.

d sin θ = nλ

d

grating spacingm

θ

angle of the maximum°

n

order of the maximum

λ

wavelengthm

Q = It

Q

chargeC

I

currentA

t

times

I = Anvq

I

currentA

A

cross-sectional aream²

n

number density of charge carriersm⁻³

v

drift speedm s⁻¹

q

charge on each carrierC

V = W / Q

V

potential differenceV

W

work doneJ

Q

chargeC

P = VI = I²R = V²/R

P

powerW

V

potential differenceV

I

currentA

R

resistanceΩ

V = IR

RearrangedR = V / I · I = V / R

V

potential differenceV

I

currentA

R

resistanceΩ

R = ρL / A

R

resistanceΩ

ρ

resistivityΩ m

L

lengthm

A

cross-sectional aream²

Σ I(in) = Σ I(out)

I

current at a junctionA

A consequence of conservation of charge.

Σ E = Σ IR

E

e.m.f. round a loopV

IR

potential difference across each componentV

A consequence of conservation of energy.

R = R₁ + R₂ + …

R

combined resistanceΩ

1/R = 1/R₁ + 1/R₂ + …

R

combined resistanceΩ

Vout = Vin × R₂ / (R₁ + R₂)

Vout

p.d. across R₂V

Vin

supply p.d.V

R₁, R₂

the two resistancesΩ

Not printed in the syllabus: it follows from V = IR with the same current through both. The syllabus requires the potential divider and the potentiometer to be used.

ᴬZX · A = Z + N

A

nucleon number

Z

proton number

N

number of neutrons

up, charm, top = +⅔e · down, strange, bottom = −⅓e

e

elementary chargeC

antiquark

carries the opposite charge

A proton is uud and a neutron is udd.

ω = 2π / T

ω

angular speedrad s⁻¹

T

periods

v = rω

v

linear speedm s⁻¹

r

radiusm

ω

angular speedrad s⁻¹

a = rω² = v²/r

a

centripetal accelerationm s⁻²

r

radiusm

v

linear speedm s⁻¹

F = mrω² = mv²/r

F

centripetal forceN

m

masskg

Directed towards the centre. Note that IGCSE explicitly excludes this equation; A Level requires it.

F = Gm₁m₂ / r²

F

gravitational forceN

G

gravitational constantN m² kg⁻²

m₁, m₂

the two point masseskg

r

separationm

g = GM / r²

g

field strengthN kg⁻¹

M

mass of the point masskg

r

distance from itm

The syllabus asks you to derive it from Newton's law of gravitation.

φ = −GM / r

φ

gravitational potentialJ kg⁻¹

M

masskg

r

distancem

Negative because the zero of potential is at infinity.

EP = −GMm / r

EP

potential energy of the pairJ

M, m

the two point masseskg

T / K = θ / °C + 273.15

T

thermodynamic temperatureK

θ

temperature°C

273.15 here, where IGCSE uses 273.

E = mcΔθ

E

energy suppliedJ

m

masskg

c

specific heat capacityJ kg⁻¹ K⁻¹

Δθ

temperature changeK or °C

E = mL

E

energy suppliedJ

m

mass changing statekg

L

specific latent heatJ kg⁻¹

Not examined at IGCSE; it appears here.

pV = nRT

p

pressurePa

V

volumem³

n

amount of substancemol

R

molar gas constantJ K⁻¹ mol⁻¹

T

thermodynamic temperatureK

pV = NkT

N

number of molecules

k

Boltzmann constantJ K⁻¹

k = R / NA

k

Boltzmann constantJ K⁻¹

R

molar gas constantJ K⁻¹ mol⁻¹

NA

Avogadro constantmol⁻¹

p = ⅓ (Nm/V) <c²>

N

number of molecules

m

mass of one moleculekg

V

volumem³

<c²>

mean-square speedm² s⁻²

½m<c²> = (3/2)kT

k

Boltzmann constantJ K⁻¹

T

thermodynamic temperatureK

Deduced by comparing pV = ⅓Nm<c²> with pV = NkT.

W = pΔV

W

work doneJ

p

pressurePa

ΔV

change in volumem³

At constant pressure.

ΔU = q + W

ΔU

increase in internal energyJ

q

energy supplied by heatingJ

W

work done on the systemJ

a = −ω²x

a

accelerationm s⁻²

ω

angular frequencyrad s⁻¹

x

displacement from equilibriumm

x = x₀ sin ωt

x₀

amplitudem

t

times

A solution of a = −ω²x.

v = v₀ cos ωt · v = ±ω√(x₀² − x²)

v₀

maximum speed, = ωx₀m s⁻¹

x₀

amplitudem

E = ½mω²x₀²

E

total energyJ

m

masskg

ω

angular frequencyrad s⁻¹

x₀

amplitudem

F = qE

F

forceN

q

chargeC

E

electric field strengthN C⁻¹, V m⁻¹

E = V / d

E

field strengthV m⁻¹

V

potential differenceV

d

plate separationm

F = Q₁Q₂ / (4πε₀r²)

F

forceN

Q₁, Q₂

the two point chargesC

ε₀

permittivity of free spaceF m⁻¹

r

separationm

In free space.

E = Q / (4πε₀r²)

E

field strengthV m⁻¹

Q

point chargeC

V = Q / (4πε₀r)

V

electric potentialV

Q

point chargeC

EP = Qq / (4πε₀r)

EP

potential energy of the pairJ

Q, q

the two point chargesC

E = − dV / dx

E

field strengthV m⁻¹

dV/dx

potential gradientV m⁻¹

C = Q / V

C

capacitanceF

Q

chargeC

V

potential differenceV

1/C = 1/C₁ + 1/C₂ + …

C

combined capacitanceF

The opposite way round from resistors — the syllabus asks you to derive both from C = Q/V.

C = C₁ + C₂ + …

C

combined capacitanceF

W = ½QV = ½CV²

W

energy storedJ

Q

chargeC

V

potential differenceV

C

capacitanceF

The area under a charge–p.d. graph.

τ = RC

τ

time constants

R

resistanceΩ

C

capacitanceF

x = x₀ e^(−t/RC)

x

current, charge or potential difference

x₀

its initial value

t

times

F = BIL sin θ

F

forceN

B

magnetic flux densityT

I

currentA

L

length in the fieldm

θ

angle between the conductor and the field°

Direction from Fleming's left-hand rule.

F = BQv sin θ

F

forceN

B

magnetic flux densityT

Q

chargeC

v

speedm s⁻¹

VH = BI / (ntq)

VH

Hall voltageV

n

number density of charge carriersm⁻³

t

thicknessm

q

charge on each carrierC

The syllabus asks you to derive it as well as use it.

Φ = BA

Φ

magnetic fluxWb

B

flux density normal to the areaT

A

aream²

E = − d(NΦ) / dt

E

induced e.m.f.V

NΦ

flux linkageWb

The magnitude is the rate of change of flux linkage; the minus sign is Lenz's law.

x = x₀ sin ωt

x

current or voltage at time t

x₀

peak value

ω

angular frequencyrad s⁻¹

Ir.m.s. = I₀/√2 · Vr.m.s. = V₀/√2

I₀, V₀

peak current and voltageA, V

For a sinusoidal alternating current only.

mean power = ½ × maximum power

P

powerW

For a sinusoidally alternating current.

E = hf

RearrangedE = hc / λ

E

photon energyJ

h

Planck constantJ s

f

frequencyHz

p = E / c

p

momentumkg m s⁻¹

E

photon energyJ

c

speed of lightm s⁻¹

hf = Φ + ½mv²max

hf

photon energyJ

Φ

work functionJ

½mv²max

maximum kinetic energy of the electronJ

λ = h / p

λ

de Broglie wavelengthm

h

Planck constantJ s

p

momentumkg m s⁻¹

hf = E₁ − E₂

E₁, E₂

the two energy levelsJ

f

frequency of the emitted photonHz

E = mc²

E

energyJ

m

masskg

c

speed of light in free spacem s⁻¹

E = c²Δm

Δm

mass defectkg

E

energy releasedJ

A = λN

A

activityBq

λ

decay constants⁻¹

N

number of undecayed nuclei

λ = 0.693 / t½

λ

decay constants⁻¹

t½

half-lifes

x = x₀ e^(−λt)

x

activity, number of nuclei or count rate

x₀

its initial value

Z = ρc

Z

specific acoustic impedancekg m⁻² s⁻¹

ρ

density of the mediumkg m⁻³

c

speed of sound in the mediumm s⁻¹

IR / I₀ = (Z₁ − Z₂)² / (Z₁ + Z₂)²

IR

reflected intensityW m⁻²

I₀

incident intensityW m⁻²

Z₁, Z₂

impedances either side of the boundary

I = I₀ e^(−μx)

μ

linear attenuation coefficientm⁻¹

x

distance travelled in the mediumm

I = I₀ e^(−μx)

I

transmitted intensityW m⁻²

I₀

incident intensityW m⁻²

μ

linear attenuation coefficientm⁻¹

x

thicknessm

F = L / (4πd²)

F

radiant flux intensityW m⁻²

L

luminosity of the sourceW

d

distancem

λmax ∝ 1 / T

λmax

peak wavelengthm

T

surface temperatureK

L = 4πσr²T⁴

L

luminosityW

σ

Stefan–Boltzmann constantW m⁻² K⁻⁴

r

stellar radiusm

T

surface temperatureK

Δλ/λ ≈ Δf/f ≈ v/c

Δλ

change in wavelengthm

v

speed of recessionm s⁻¹

c

speed of lightm s⁻¹

For v much less than c.

v ≈ H₀d

v

speed of recessionm s⁻¹

H₀

Hubble constants⁻¹

d

distance to the galaxym